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Calculate the flow through a plane

  1. Dec 25, 2013 #1
    1. The problem statement, all variables and given/known data
    Let ##\vec{F}(x,y,z)## be defined as ##\vec{F}(x,y,z)=(-x+y^2+z^3,xe^y,-xze^y+z+1)## and plane ##\Sigma## defined with parameterization ##\vec{r}(u,v)=((1-u^2-v^2)(u^2+v^2)(sinh^3u+v)+u,v+(1-u^2-v^2)sinhu,-u^2-v^2+1)## where u,v are elements of a circle in point (0,0) with a radius 1. Orientation of ##\Sigma## is regulated in accordance with parametrization.

    Calculate the flow of ##\vec{F}## through plane ##\Sigma##.
    2. Relevant equations



    3. The attempt at a solution
    Hi there!

    I have a problem... I have no idea how to start on this problem? I know that I can replace the x,y,z in ##\vec{F}## with u and v from parameterization ##\vec{r}## but this would give me some massive never ending calculations.

    Any other way to do this?
     
  2. jcsd
  3. Dec 25, 2013 #2

    Astronuc

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    Staff: Mentor

    Can one think of a relevant equation that describes a flow based on the flux through an area?

    What is [itex]\vec{F}[/itex]?

    Is ##\vec{r}(u,v)=((1-u^2-v^2)(u^2+v^2)(sinh^3(u+v))+u,v+(1-u^2-v^2) sinh u,-u^2-v^2+1)## correct?
     
  4. Dec 25, 2013 #3
    You're probably talking about Gaussian law... but, to be honest with you, I don't know when am I allowed to use it...

    Am... there is one little mistake (sorry for that).. it's ##\vec{r}(u,v)=((1-u^2-v^2)(u^2+v^2)(sinh^3(u)+v)+u,v+(1-u^2-v^2) sinh (u),-u^2-v^2+1)##
     
  5. Dec 25, 2013 #4

    HallsofIvy

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    Staff Emeritus
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    First, is that really a plane and not a curved surface? I'm not saying it isn't because I haven't checked, but a plane can be written in terms of linear[/b ] functions of the parameters but doesn't have to be . In any case, if it is a plane then the normal vector has constant direction.

    The flow of [itex]\vec{F}(x,y,z)= f(x, y, z)\vec{i}+ g(x,y,z)\vec{j}+ h(x,y,z)\vec{k}[/itex] through a surface is the integral of the dot product of [itex]\vec{F}[/itex] and the unit normal vector to the surface at each point.

    The "normal vector" to a surface given by [itex]\vec{r}(u,v)[/itex] is given by the cross product of [itex]\vec{r_u}\times \vec{r_v}. The unit normal is, of course, that normal vector divided by its own length.
     
  6. Dec 25, 2013 #5
    Aaam, I'm not sure if that is a plane or a curved surface. If I would have to guess i'd say it's a curved surface, however i am not really even sure how one can check that?

    Ok, but I assume I have to change ##\vec{F}(x,y,z)## into ##\vec{F}(u,v)## or not?
     
  7. Dec 26, 2013 #6
    ##\vec{r}(u,v)=((1-u^2-v^2)(u^2+v^2)(sinh^3(u)+v)+u,v+(1-u^2-v^2) sinh (u),-u^2-v^2+1)##

    Than ##\vec{r_u}=(-2u(u^2+v^2-1)(sinh^3(u)+v)-2u(u^2+v^2)(sinh^3(u))+v)-3(u^2+v^2-1)(u^2+v^2)sinh^2(u)cosh(u)+1,-(u^2+v^2-1)cosh(u)-2usinh(u),-2u)##

    and

    ##\vec{r_v}=(-u^4+u^2(1-6v^2)-2v(2u^2+2v^2-1)sinh^3(u)-5v^4+3v^2,1-2vsinh(u),-2v)##

    is there anything I can do before calculating the normal vector ##\vec{r_u}\times \vec{r_v}## ?? Please say yes :D
     
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